q = 350 − 5p

(a) Enter the equation for revenue R as a function of price p.
R =

(b) What is the (instantaneous) rate of change of revenue with respect to price when the price is $6?

(c) At what price is the derivative of revenue with respect to price exactly zero?

I only need help with B. I have already figured out A and C

(a) We can find the revenue R by multiplying the price p by the quantity q:

R = p * q

Since q = 350 - 5p, we can substitute this value in for q:

R = p * (350 - 5p)
R = 350p - 5p^2

So, the equation for revenue R as a function of price p is R = 350p - 5p^2.

(b) To find the instantaneous rate of change of revenue with respect to price, we need to find the derivative of the revenue function R with respect to p and evaluate it at a price of $6. Let's find the derivative:

R' = dR/dp = d/dp (350p - 5p^2)
= 350 - 10p

Now, let's evaluate this derivative at p = 6:

R'(6) = 350 - 10(6)
= 350 - 60
= 290

Therefore, the (instantaneous) rate of change of revenue with respect to price when the price is $6 is $290.

(c) To find the price where the derivative of revenue with respect to price is exactly zero, we need to set R' equal to zero and solve for p:

350 - 10p = 0
10p = 350
p = 35

Hence, the price at which the derivative of revenue with respect to price is exactly zero is $35.

(a) To find the revenue R as a function of price p, we need to multiply the price by the quantity sold. The quantity sold is given by q, so the revenue formula is:

R = p * q

Substituting q = 350 - 5p, we have:

R = p * (350 - 5p)

Simplifying this equation gives us the revenue function:

R = 350p - 5p^2

(b) To find the instantaneous rate of change of revenue with respect to price, we need to take the derivative of the revenue function with respect to p. In this case, the derivative represents the rate of change of revenue with respect to price.

So, let's take the derivative of R with respect to p:

dR/dp = 350 - 10p

To find the instantaneous rate of change when the price is $6, substitute p = 6 into the derivative equation:

dR/dp = 350 - 10(6)

Simplifying further:

dR/dp = 350 - 60

Thus, the instantaneous rate of change of revenue with respect to price when the price is $6 is:

dR/dp = 290

(c) To find the price at which the derivative of revenue with respect to price is exactly zero, we need to solve the derivative equation dR/dp = 0.

So, let's set the derivative equal to zero and solve for p:

350 - 10p = 0

-10p = -350

p = 35

Therefore, at a price of $35, the derivative of revenue with respect to price is exactly zero.

(a) To find the equation for revenue R as a function of price p, we use the formula for revenue, which is the product of price and quantity. In this case, the quantity is not given explicitly, but we can relate it to the price using the equation q = 350 - 5p. So, we have:

R = p * q

Substituting the value of q from the given equation, we get:

R = p * (350 - 5p)

Simplifying this equation gives us the equation for revenue R as a function of price p.

R = 350p - 5p^2

(b) To find the instantaneous rate of change of revenue with respect to price when the price is $6, we need to find the derivative of the revenue function with respect to price and evaluate it at p = 6.

R = 350p - 5p^2

Taking the derivative with respect to p, we have:

dR/dp = 350 - 10p

Now, substitute p = 6 into the derivative:

dR/dp = 350 - 10(6)
= 350 - 60
= 290

The instantaneous rate of change of revenue with respect to price when the price is $6 is 290.

(c) To find the price at which the derivative of revenue with respect to price is exactly zero, we need to find the value of p for which dR/dp = 0.

Setting the derivative equal to zero, we have:

350 - 10p = 0

Subtracting 350 from both sides, we get:

-10p = -350

Dividing by -10, we find:

p = 35

So, the price at which the derivative of revenue with respect to price is exactly zero is $35.